python 计算目标检测结果与实际的IOU

def computeIOU(local_xml_info_list, need_check_info):
    xmin1, ymin1, xmax1, ymax1 = local_xml_info_list
    xmin2, ymin2, xmax2, ymax2 = need_check_info

    # box1 = [xmin1, ymin1, xmax1, ymax1] #左上角的坐标与右下角的坐标
    # box2 = [xmin2, ymin2, xmax2, ymax2]
    # # iou_ratio--交并比
    # width1 = abs(box1[2] - box1[0])
    # height1 = abs(box1[1] - box1[3]) # 这里y1-y2是因为一般情况y1>y2，为了方便采用绝对值
    # width2 = abs(box2[2] - box2[0])
    # height2 = abs(box2[1] - box2[3])
    # x_max = max(box1[0],box1[2],box2[0],box2[2])
    # y_max = max(box1[1],box1[3],box2[1],box2[3])
    # x_min = min(box1[0],box1[2],box2[0],box2[2])
    # y_min = min(box1[1],box1[3],box2[1],box2[3])
    # iou_width = x_min + width1 + width2 - x_max
    # iou_height = y_min + height1 + height2 - y_max
    # if iou_width <= 0 or iou_height <= 0:
    #     iou_ratio = 0
    # else:
    #     iou_area = iou_width * iou_height # 交集的面积
    #     box1_area = width1 * height1
    #     box2_area = width2 * height2
    #     iou_ratio = iou_area / (box1_area + box2_area - iou_area) # 并集的面积
    # return iou_ratio

    # 求交集部分左上角的点
    xmin = max(xmin1, xmin2)
    ymin = max(ymin1, ymin2)
    # 求交集部分右下角的点
    xmax = min(xmax1, xmax2)
    ymax = min(ymax1, ymax2)
    # 计算输入的两个矩形的面积
    s1 = (xmax1 - xmin1) * (ymax1 - ymin1)
    s2 = (xmax2 - xmin2) * (ymax2 - ymin2)

    # 计算总面积
    s = s1 + s2
    # 计算交集
    inter_area = (xmax - xmin) * (ymax - ymin)
    iou = inter_area / (s - inter_area)

转载及参考：

https://blog.csdn.net/lmxt520/article/details/98473024

https://blog.csdn.net/mndlgzzd/article/details/107604446